# Laplace operator and tensor calculus:

I'm studying Tensor calculus and I found this interesting problem:

Show that: $$\Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)$$

Here's some attempts, hope it helps, even I find them useless!

Well, we know that: $$\Delta F=\nabla\cdot \nabla F$$ And : $$\nabla \cdot \mathbf{V}=\nabla_iv ^i$$Using it : $$\Delta F=\nabla_i (g^{ik}\partial_kF)$$

That's the only advance I've made till now, I'm thinking about a property but I'm not that much certain about its validity here.

$$\Delta F=g^{ik}\nabla_i(\partial_k F)$$

Being true or false I think it's not useful to derive this formula.

• Since this is purely a math question, it would be more appropriate on Mathematics. – G. Smith Jul 25 at 19:10
• The divergence of a vector field $\mathbf v$ is only given by $\nabla_i v^i$ in Cartesian coordinates. In general it will involve Christoffel symbols (though after some work, you can express it in terms of $|g|$). You might be interested in this post from MathSE. – J. Murray Jul 25 at 21:18

Not from the first principles, but based on physical intuition proof looks as follows. Consider the action for scalar field: $$S = \int d^D x \ \sqrt{g} g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi$$ This is the only viable GR covariant expression for the action of scalar field without free indices, and $$d^D x \sqrt{g}$$ is an invariant volume element. Integrating this expression by parts, one gets: $$S = -\int d^D x \ \phi \partial_\mu (\sqrt{g} g^{\mu \nu} \partial_\nu \phi) = - \int d^D x \ \sqrt{g} \phi \frac{1}{\sqrt{g}}\partial_\mu (\sqrt{g} g^{\mu \nu} \partial_\nu \phi) = -\int d^D x \ \phi \Delta \phi$$ Where we have assumed that boundary terms vanis, and recovered in the last equality the invariant volume element.

Well, the Laplace operator is a composite operator:

$$\Delta F = div\ grad\ F = \nabla\cdot\nabla F$$

and as you wrote

$$(grad\ F)^r = (\nabla F)^r = \frac{\partial F}{\partial x^k}\,g^{rk} = V^r$$

You obtain the divergence by contraction of the derivation operator $$\nabla$$ and we emphasize that the contraction has to be performed on the covariant derivative:

$$div\ \boldsymbol{V} = \nabla_iV^i = V^i_{\phantom{i};\,i}= \frac{\partial V^i}{\partial x^i} + V^r\; \Gamma^i_{ir}$$

By use of a property of the levi-Civita connection coefficients

$$\Gamma^i_{ki} = \frac{1}{2} g^{ij} \frac{\partial g_{ij}}{\partial x^k} = \frac{1}{2g} \frac{\partial g}{\partial x^k} = \frac{\partial \,log \sqrt{|g|}}{\partial x^k}$$

you can write further

$$div\ \boldsymbol{V} = \nabla_iV^i = V^i_{\phantom{i};\,i}= \frac{\partial V^i}{\partial x^i} + V^r\; \Gamma^i_{ir} = \frac{\partial V^r}{\partial x^r} + V^r\; \frac{\partial \,log \sqrt{|g|}}{\partial x^r} = \frac{1}{\sqrt{|g|}}\; \frac{\partial}{\partial x^r} (\sqrt{|g|}\; V^r)$$

Finally, substituting $$V^r$$ gives the desired result:

$$\Delta F = div\ grad\ F = \frac{1}{\sqrt{|g|}}\; \frac{\partial}{\partial x^r} (\sqrt{|g|}\; \frac{\partial F}{\partial x^k}\,g^{rk} )$$

Here's a quick derivation of this problem:

As you said: $$\Delta F= \nabla\ .\nabla F$$ And always using your steps: $$\nabla \ .\ F=\nabla_iv^i$$ And for those who don't know why he involved the "($$g^{ik}\partial_k F$$)" well that's the contravariant components of the gradient operator. \begin{align} \Delta F&= \nabla_iv^i\\ &=\nabla_i\left(g^{ik}\partial_kF\right)\\ &=g^{ik}\nabla_i\left(\partial_k F\right) \end{align}

Recall: $$\nabla_i(\partial_k F)=\partial_{ik}F-\Gamma_{ik}^l\partial_lF$$

Hence: $$\Delta F=g^{ik}(\partial_{ik}F-\Gamma_{ik}^l\partial_lF)$$

Another recall: :) $$\nabla\ .\ \mathbf{V}=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(v^i \sqrt{\vert g\vert}\right)\quad{(1)}$$

Involving the contravariant components of $$\mathbf{grad}F$$ in $$(1)$$ we got the following: $$\bbox[silver,5px,border:2px solid teal] {\Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)}$$ and that's true, because when $$g^{ik}=\delta^{ik}$$ we get the classic expression of the Laplacien operator: $$\Delta \ F=\partial_{kk}F.$$